Intercepts on the Axes made by a Circle

We will learn how to find the intercepts on the axes made by a circle.

The lengths of intercepts made by the circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 with X and Y axes are 2\(\mathrm{\sqrt{g^{2} - c}}\) and 2\(\mathrm{\sqrt{f^{2} - c}}\) respectively.

Proof:

Let the given equation of the circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 ………. (1)

Clearly, the centre of the circle is c (-g, -f) and the radius = \(\mathrm{\sqrt{g^{2} + f^{2}- c}}\)

Let AB be the intercept made by the given circle on x-axe. Since on x-axis, y = 0. Therefore, x-coordinates of the points A and B are the roots of the equation x\(^{2}\) + 2gx + c = 0.

Let x\(_{1}\) and x\(_{2}\) be the x-coordinates of the points A and B respectively. Then, x\(_{1}\) and x\(_{2}\) also the roots of the equation x\(^{2}\) + 2gx + c = 0.

Therefore, x\(_{1}\) + x\(_{2}\) = - 2g and x\(_{1}\)x\(_{2}\) = c

Clearly the intercept on x-axis = AB

                                          = x\(_{2}\) - x\(_{1}\) = \(\mathrm{\sqrt{(x_{2} - x_{1})^{2}}}\)

                                          = \(\mathrm{\sqrt{(x_{2} + x_{1})^{2} - 4x_{1}x_{2}}}\)

                                          = \(\mathrm{\sqrt{4g^{2} - 4c}}\)

                                          = 2\(\mathrm{\sqrt{g^{2} - c}}\)

Therefore, the intercept made by the circle (1) on the x-axis = 2\(\mathrm{\sqrt{g^{2} - c}}\)

Again,

Let DE be the intercept made by the given circle on y-axe. Since on y-axis, x = 0. Therefore, y-coordinates of the points D and E are the roots of the equation y\(^{2}\) + 2fy + c = 0.

Let y\(_{1}\) and y\(_{2}\) be the x-coordinates of the points D and E respectively. Then, y\(_{1}\) and y\(_{2}\) also the roots of the equation y\(^{2}\) + 2fy + c = 0

Therefore, y\(_{1}\) + y\(_{2}\) = - 2f and y\(_{1}\)y\(_{2}\) = c

Clearly the intercept on y-axis = DE

                                          = y\(_{2}\) - y\(_{1}\) = \(\mathrm{\sqrt{(y_{2} - y_{1})^{2}}}\)

                                          = \(\mathrm{\sqrt{(y_{2} + y_{1})^{2} – 4y_{1}y_{2}}}\)

                                          = \(\mathrm{\sqrt{4f^{2} - 4c}}\)

                                          = 2\(\mathrm{\sqrt{f^{2} - c}}\)

Therefore, the intercept made by the circle (1) on the y-axis = 2\(\mathrm{\sqrt{f^{2} - c}}\)

Solved examples to find the intercepts made by a given circle on the co-ordinate axes:

1. Find the length of the x-intercept and y-intercept made by the circle x\(^{2}\) + y\(^{2}\) - 4x -6y - 5 = 0 with the co-ordinate axes.

Solution:

Given equation of the circle is x\(^{2}\) + y\(^{2}\) - 4x -6y - 5 = 0.

Now comparing the given equation with the general equation of the circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0, we get g = -2 and f = -3 and c = -5

Therefore, length of the x-intercept = 2\(\mathrm{\sqrt{g^{2} - c}}\) = 2\(\mathrm{\sqrt{4 - (-5) }}\) = 2√9 = 6.

The length of the y-intercept = 2\(\mathrm{\sqrt{f^{2} - c}}\) = 2\(\mathrm{\sqrt{9 - (-5) }}\) = 2√14.


2. Find the equation of a circle which touches the y-axis at a distance -3 from the origin and cuts an intercept of 8 units with the positive direction of x-axis.

Solution:

Let the equation of the circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 …………….. (i)

According to the problem, the equation (i) touches the y-axis

Therefore, c = f\(^{2}\) ………………… (ii)

Again, the point (0, -3) lies on the circle (i).

Therefore, putting the value of x = 0 and y = -3 in (i) we get,

9 - 6f + c = 0 …………………… (iii)

From (ii) and (iii), we get 9 - 6f + f\(^{2}\) = 0 ⇒ (f - 3)\(^{2}\) = 0 ⇒ f - 3 = 0 ⇒ f = 3

Now putting f = 3 in (i) we get, c = 9

Again, according to the problem the equation of the circle (i) cuts an intercept of 8 units with the positive direction of x-axis.

Therefore,

2\(\mathrm{\sqrt{g^{2} - c}}\) = 8

⇒ 2\(\mathrm{\sqrt{g^{2} - 9}}\) = 8

⇒ \(\mathrm{\sqrt{g^{2} - 9}}\) = 4

⇒ g\(^{2}\) - 9 = 16, [Squaring both sides]

⇒ g\(^{2}\) = 16 + 9

⇒ g\(^{2}\) = 25

⇒ g = ±5.

Hence, the required equation of the circle is x^2 + y^2 ± 10x + 6y + 9 = 0.

 The Circle




11 and 12 Grade Math 

From Intercepts on the Axes made by a Circle to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Adding 1-Digit Number | Understand the Concept one Digit Number

    Sep 17, 24 02:25 AM

    Add by Counting Forward
    Understand the concept of adding 1-digit number with the help of objects as well as numbers.

    Read More

  2. Counting Before, After and Between Numbers up to 10 | Number Counting

    Sep 17, 24 01:47 AM

    Before After Between
    Counting before, after and between numbers up to 10 improves the child’s counting skills.

    Read More

  3. Worksheet on Three-digit Numbers | Write the Missing Numbers | Pattern

    Sep 17, 24 12:10 AM

    Reading 3-digit Numbers
    Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures…

    Read More

  4. Arranging Numbers | Ascending Order | Descending Order |Compare Digits

    Sep 16, 24 11:24 PM

    Arranging Numbers
    We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…

    Read More

  5. Worksheet on Tens and Ones | Math Place Value |Tens and Ones Questions

    Sep 16, 24 02:40 PM

    Tens and Ones
    In math place value the worksheet on tens and ones questions are given below so that students can do enough practice which will help the kids to learn further numbers.

    Read More